# Trying to understand relativity of lengths and times

• myoho.renge.kyo
In summary: L has ends A and B with clocks attached to it. In the moving frame, the light travels from A to B a distance L in time TB - TA and from B to A a distance L in time T‘A - TB. In the rest frame, the light travels from A to B a distance L in time TB - TA and from B to A a distance L in time T‘A - TB. The clocks at A and B were adjusted before the stick began to move in the rest frame in order to synchronize with the clocks in the rest frame.
myoho.renge.kyo
let K' = a stationary rigid rod. let its length be L (the length between the two ends A and B of K' using a measuring-rod). the axis of K' is lying along the x-axis of a stationary system of coordinates K.

let K be provided with clocks which synchronize.

let a constant speed v in the direction of increasing x be imparted to K'.

let rAB = the distance (as measured by the measuring-rod already employed to measure L) between two points on the x-axis of K where the two ends A and B of K’ are located at a definite time.

let clocks be placed at the two ends A and B of K’, and let these clocks synchronize with the clocks of K

let a ray of light depart from A at tA in the direction of B, and let the ray of light be reflected at B at tB and reach A again at t’A.

time here denotes time of K and also position of hands of the moving clock at the place under discussion.

observers moving with K' would declare that (tB - tA) = rAB/(c - v) is not equal to (t’A - tB) = rAB/(c + v).

while observers in K would declare that (tB - tA) = rAB/(c - v) = (t’A - tB) = rAB/(c + v).

if L = 299,792,458 meters, and v = 0.5*c, what is the value of (tB - tA) and (t’A - tB) using the lorentz transformations if the observers are moving with K'?

and if L = 299,792,458 meters, and v = 0.5*c, what is the value of (tB - tA) and (t’A - tB) using the lorentz transformations if the observers are in K?

thanks. i think the answer is going to help me understand the lorentz transformations and how to apply them in order to understand relativity of lengths and times. thanks again.

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I find it extremely difficult to understand your posts. Let me try my own version of what you might be getting at:

A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

A beam of light flashes at A, travels to B, then is reflected back to A.

Realize that different observers will measure different times and locations for these events (the flash, the reflection, the return).

In the frame of the stick (let's call that the "moving frame"), the light travels a distance L in time T. c = L/T.

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the light takes time T' to get from A to B. The distance covered is L' + vT', where L' is the length of the stick according to the rest frame and vT' is the distance that end B moves in that time according to the rest frame.

Now, the time for the round trip of the light--according to the moving frame--is just 2T. That time will be recorded on the clock at point A.

Now, according to the rest frame, the time for round trip of the light will be 2T/sqrt(1 - v^2/c^2).

With this as a start, can you rephrase your question?

Doc Al said:
I find it extremely difficult to understand your posts. Let me try my own version of what you might be getting at:

A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

A beam of light flashes at A, travels to B, then is reflected back to A.

Realize that different observers will measure different times and locations for these events (the flash, the reflection, the return).

In the frame of the stick (let's call that the "moving frame"), the light travels a distance L in time T. c = L/T.

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the light takes time T' to get from A to B. The distance covered is L' + vT', where L' is the length of the stick according to the rest frame and vT' is the distance that end B moves in that time according to the rest frame.

Now, the time for the round trip of the light--according to the moving frame--is just 2T. That time will be recorded on the clock at point A.

Now, according to the rest frame, the time for round trip of the light will be 2T/sqrt(1 - v^2/c^2).

With this as a start, can you rephrase your question?

A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

A beam of light flashes at A at TA (the position of the hands of the clock at A), travels to B at TB (the position of the hands of the clock at B), then is reflected back to A at T‘A (the position of the hands of the clock at A).

In the frame of the stick (let's call that the "moving frame"), the light travels from A to B a distance L in time TB - TA. c = L/(TB - TA). And in the moving frame, the light travels from B to A a distance L in time T‘A - TB. c = L/(T’A - TB).

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the beam of light flashes at A at TA, travels to B at TB, then is reflected back to A at T’A.

Realize that the clocks at A and B were adjusted before the stick began to move at speed v in the rest frame in order that they synchronize with the clocks in the rest frame when the stick is then seen to move at speed v in the rest frame.

In the rest frame, the light travels from A to B a distance L’ (where L' is the length of the stick according to the rest frame) in time TB - TA. c = L’/(TB - TA) + v. And in the rest frame, the light travels from B to A in time T’A - TB. c = L’/(T’A - TB) - v. TB - TA = T’A - TB.

In the moving frame, c = L’/(TB - TA) + v and c = L’/(T’A - TB) - v. (TB - TA) is not equal to (T’A - TB).

what is the value of (TB - TA), (T'A - TB), and L' in the rest frame and the moving frame respectively if L = 299,792,458 meters? thanks.

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myoho.renge.kyo said:
A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

A beam of light flashes at A at TA (the position of the hands of the clock at A), travels to B at TB (the position of the hands of the clock at B), then is reflected back to A at T‘A (the position of the hands of the clock at A).

In the frame of the stick (let's call that the "moving frame"), the light travels from A to B a distance L in time TB - TA. c = L/(TB - TA). And in the moving frame, the light travels from B to A a distance L in time T‘A - TB. c = L/(T’A - TB).

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the beam of light flashes at A at TA, travels to B at TB, then is reflected back to A at T’A.

Realize that the clocks at A and B were adjusted before the stick began to move at speed v in the rest frame in order that they synchronize with the clocks in the rest frame when the stick is then seen to move at speed v in the rest frame.
You mean the clocks A and B are now in sync in the frame where the stick moves at velocity v, and out-of-sync in the rest frame of the stick? But do you also mean that the rates of the clocks at A and B are adjusted so that they appear to tick at the correct rate in the "rest frame" where the stick is moving at speed v? In other words, the two clocks have been artificially sped up to tick at $$1/\sqrt{1 - v^2/c^2}$$ the normal rate in their own rest frame?
myoho.renge.kyo said:
In the rest frame, the light travels from A to B a distance L’ (where L' is the length of the stick according to the rest frame) in time TB - TA. c = L’/(TB - TA) + v. And in the rest frame, the light travels from B to A in time T’A - TB. c = L’/(T’A - TB) - v. TB - TA = T’A - TB.
"Rest frame" refers to the frame where the stick is moving, right? Your post would be more clear if instead of "rest frame" and "moving frame" you specified whose rest frame you were talking about, as in "the rest frame of the stick" or "the rest frame of the observer who sees the stick moving at velocity v". But assuming you mean the latter, then if the clocks at A and B have been adjusted so that they are not only synchronized in the observer's rest frame, but also ticking at the correct rate in the observer's frame, then these equations will be correct, because in that case (TB - TA) will be the time the observer measures between the light passing A and the light passing B, and during that time the ruler will have moved forward by a distance of v*(TB - TA), so the total distance the light will have covered between those events is L' + v*(TB - TA) in this frame, and the total time between the events is (TB - TA) in this frame, and distance/time in the observer's frame must be c. Likewise, on the way back the light will only have to cover a distance of L' - v*(T'A - TB) because A is moving towards the point where the light was reflected, so your second equation is also right.
myoho.renge.kyo said:
In the moving frame, c = L’/(TB - TA) + v and c = L’/(T’A - TB) - v.
Why would you say "in the moving frame" when you are using clocks synchronized in the observer's rest frame (and, if I'm understanding right, also running at the correct rate in the observer's rest frame), and the length L' seen in the observer's rest frame? When you say "in frame x" you normally mean "using frame x's definition of the distance between the events and frame x's definition of the time between them".
myoho.renge.kyo said:
(TB - TA) is not equal to (T’A - TB).
If the clocks at A and B are synchronized in the frame of the observer who sees the ruler moving at velocity v, that's correct.
myoho.renge.kyo said:
what is the value of (TB - TA), (T'A - TB), and L' in the rest frame and the moving frame respectively if L = 299,792,458 meters? thanks.
Assuming (TB - TA) is simply the time for the light to go from A to B in the frame of the observer who sees the ruler moving at v, and L' is the length of the ruler in that observer's frame, then L' is easy to find, it's just $$299,792,458*\sqrt{1 - v^2/c^2}$$ due to length contraction. To find (TB - TA) and (T'A - TB) you'd have to also know v, but it would just be a matter of solving the equations you already gave, c = L'/(TB - TA) + v (which can be rearranged to give (TB - TA) = L'/(c - v)) and c = L'/(T'A - TB) - v (which can be rearranged to give (T'A - TB) = L'/(c + v)).

myoho.renge.kyo said:
A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

A beam of light flashes at A at TA (the position of the hands of the clock at A), travels to B at TB (the position of the hands of the clock at B), then is reflected back to A at T‘A (the position of the hands of the clock at A).

In the frame of the stick (let's call that the "moving frame"), the light travels from A to B a distance L in time TB - TA. c = L/(TB - TA). And in the moving frame, the light travels from B to A a distance L in time T‘A - TB. c = L/(T’A - TB).
So far, so good.

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the beam of light flashes at A at TA, travels to B at TB, then is reflected back to A at T’A.

Realize that the clocks at A and B were adjusted before the stick began to move at speed v in the rest frame in order that they synchronize with the clocks in the rest frame when the stick is then seen to move at speed v in the rest frame.
ah... here's a problem. By starting with the stick "at rest" and then having it move at speed "v" you are introducing an acceleration and its associated complications. In order to preserve the synchonization of clocks A and B as the stick is accelerated, you would have to accelerate the ends of the stick (and all points in between) uniformly with respect to the "rest frame". The stick will break (or at least be forced to stretch) under the stress of that operation--it will no longer have a proper length L. Even if you succeed in getting the stick to move at speed v without disintegrating, the clocks will still be running slowly compared to clocks in the rest frame.

I strongly advise not introducing such accelerations when first trying to sort out the basics of special relativity. Instead, imagine the stick to have been always in motion at speed v with respect to the "rest frame". Pose your questions with respect to that model and it'll be much easier to see how the Lorentz transformations work.

Doc Al said:
ah... here's a problem. By starting with the stick "at rest" and then having it move at speed "v" you are introducing an acceleration and its associated complications. In order to preserve the synchonization of clocks A and B as the stick is accelerated, you would have to accelerate the ends of the stick (and all points in between) uniformly with respect to the "rest frame". The stick will break (or at least be forced to stretch) under the stress of that operation--it will no longer have a proper length L. Even if you succeed in getting the stick to move at speed v without disintegrating, the clocks will still be running slowly compared to clocks in the rest frame.

i understand. let me try it this way (i am quoting A. Enstein in the process, pages 40-42 in The Principle of Relativity):

A stick, proper length L, has ends A and B.

In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), there are clocks set up on the x-axis of this "rest frame". These clocks are synchronized. L' is the distance between two points on the x-axis of this "rest frame" where the two ends A and B of the stick are located at a definite time.

There are clocks attached at the two ends A and B of the stick which synchronize with the clocks set up on the x-axis of the "rest frame", that is to say that their indications correspond at any instant to the "time of the 'rest frame'" at the places where they happened to be. These clocks are therefore, synchronous in the "rest frame".

A beam of light flashes at A at the time* TA (the position of the hands of the clock at A), travels to B at the time TB (the position of the hands of the clock at B), then is reflected back to A at the time T‘A (the position of the hands of the clock at A). The two clocks synchronize if

TB - TA = T'A - TB

for simplicity, let us assume that the moving clocks and the clocks in the "rest frame" simultaneously mark TA = 0 s.

*"time" here denotes "time of the 'rest frame'" and also "position of hands of the moving clock situated at the place under discussion."

Taking into consideration the principle of the constancy of the velocity of light we find that

TB - TA = L'/(c - v) and T'A - TB = L'/(c + v)

observers moving with the stick would find that the two clocks were not synchronous, while observers in the "rest frame" would declare the clocks to be synchronous.

using the lorentz transformations, find the value of (TB - TA) and L' if L = 299,792,458 meters and v = 0.5*c, such that the observers moving with the stick would find that the two clocks were not synchronous, while the observers in the "rest frame" would declare the clocks to be synchronous. thanks!

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myoho.renge.kyo said:
TB - TA = L'/(c - v) and T'A - TB = L'/(c + v)
These are the formulas I gave at the end of my last post.
myoho.renge.kyo said:
]observers moving with the stick would find that the two clocks were not synchronous, while observers in the "rest frame" would declare the clocks to be synchronous.
As I mentioned above, this scenario is also assuming the clocks are ticking at the correct rate in the "rest frame", which means that in the frame of the stick the clocks would appear artificially sped up by a factor of $$1/\sqrt{1 - v^2/c^2}$$.
myoho.renge.kyo said:
using the lorentz transformations, what is the value of (TB - TA), (T'A - TB), and L' in the "rest frame" and in the moving stick respectively if L = 299,792,458 meters and v = 0.5*c? thanks.
Just plug into the formulas, using the length contraction formula to find L' given L. In this case L' = 299,792,458*$$\sqrt{1 - 0.5^2}$$ meters, so TB - TA = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(1c - 0.5c) = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(299,792,458 meters/second)*(0.5) = $$(\sqrt{1 - 0.5^2})/0.5$$*(1 second) = 1.73205081 seconds. Similarly, T'A - TB = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(1c + 0.5c) = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(299,792,458 meters/second)*(1.5) = $$(\sqrt{1 - 0.5^2})/1.5$$*(1 second) = 0.577350269 seconds.

myoho.renge.kyo said:
i understand. let me try it this way ...

using the lorentz transformations, what is the value of (TB - TA), (T'A - TB), and L' in the "rest frame" and in the moving stick respectively if L = 299,792,458 meters and v = 0.5*c? thanks.

v = 0.5c
gamma = 1.1547

Rest Frame …

TB-TA = T(out) = 1 sec = 1/2(2L/c)
T'A-TB = T(back) = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters (proper length)

Moving frame …

TB-TA = T(out) = 1.732 sec = (L/gamma)[1/(c-v)]
T'A-TB = T(back) = 0.57733 sec = (L/gamma)[1/(c+v)]
L’ = 0.866 lt-sec = 259,620,268 meters = L/gamma (contracted length)

By the Lorentz Transformations ...

Per Rest Frame (by definition) ...

TB-TA = 1 sec = 1/2(2L/c)
T'A-TB = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters

Per Moving Frame (per LTs) ...

t(out) = gamma(T+vX/c2) = TB-TA

t(out) = gamma(T+vX/c2)
t(out) = (1.1547)[(1)+(0.5)(1)/(1)2]
t(out) = 1.732 sec​

t(back) = gamma(T-vX/c2) = T'A-TB

t(back) = gamma(T-vX/c2)
t(back) = (1.1547)[(1)-(0.5)(1)/(1)2]
t(back) = 0.5774 sec​

Verification ... Roundtrip

t(out)+t(back) = gamma[T(out)+T(back)]
1.732 sec + 0.5774 sec = (1.1547)[1 sec + 1sec]
2.3094 sec = 2.3094 sec​

x(out) = gamma(X-vT)

x(out) = gamma(X-vT)
x(out) = (1.1547)[(1)-(0.5c)/(1)2]
x(out) = (1.1547)[1-0.5]
x(out) = 1.732 lt-sec​

x(back) = gamma(X+vT)

x(back) = gamma(X+vT)
x(back) = (1.1547)[(1)+(0.5c)/(1)2]
x(back) = (1.1547)[1+0.5]
x(back) = 0.5774 lt-sec​

L' = x’ = (x-vt) ... calculated by the outbound leg

x’ = (x-vt)
x’ = ct-vt
x’ = t(c-v)
x’ = (1.732 sec)(1-0.5)
x’ = (1.732 sec)0.5
x’ = 0.866 lt-sec
L' = x’ = 0.866 lt-sec = L/gamma

L' = x’ = (x+vt) ... calculated by the return leg

x’ = (x+vt)
x’ = ct+vt
x’ = t(c+v)
x’ = (0.5774 sec)(1+0.5)
x’ = (0.5774 sec)1.5
x’ = 0.866 lt-sec
L' = x’ = 0.866 lt-sec = L/gamma

pess

pess5 said:
Per Rest Frame (by definition) ...

TB-TA = 1 sec = 1/2(2L/c)
T'A-TB = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters

Per Moving Frame (per LTs) ...

t(out) = gamma(T+vX/c2) = TB-TA

t(out) = gamma(T+vX/c2)
t(out) = (1.1547)[(1)+(0.5)(1)/(1)2]
t(out) = 1.732 sec​

t(back) = gamma(T-vX/c2) = T'A-TB

t(back) = gamma(T-vX/c2)
t(back) = (1.1547)[(1)-(0.5)(1)/(1)2]
t(back) = 0.5774 sec​
myoho.renge.kyo is actually defining "rest frame" and "moving frame" in the opposite way--the "moving frame" is the rest frame of the stick (where its length is 1 light-second), and the "rest frame" is the frame where the stick is moving at velocity v. Also, TB - TA and T'A - TB are defined to be the time intervals in the "rest frame" where the stick is moving, not the time in the stick's own frame. This sort of confusion is why I made this suggestion earlier:
Your post would be more clear if instead of "rest frame" and "moving frame" you specified whose rest frame you were talking about, as in "the rest frame of the stick" or "the rest frame of the observer who sees the stick moving at velocity v".
Aside from that, your numbers for the times agree with the ones I got in my last post.

pess5 said:
v = 0.5c
gamma = 1.1547

Rest Frame …

TB-TA = T(out) = 1 sec = 1/2(2L/c)
T'A-TB = T(back) = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters (proper length)

you say "Rest Frame..."

what do you mean?

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JesseM said:
myoho.renge.kyo is actually defining "rest frame" and "moving frame" in the opposite way.

the "rest frame" is the frame in which the stick is seen to move at speed v.

the "moving frame" is the stick that moves at speed v in the "rest frame".

JesseM said:
These are the formulas I gave at the end of my last post. As I mentioned above, this scenario is also assuming the clocks are ticking at the correct rate in the "rest frame", which means that in the frame of the stick the clocks would appear artificially sped up by a factor of $$1/\sqrt{1 - v^2/c^2}$$. Just plug into the formulas, using the length contraction formula to find L' given L. In this case L' = 299,792,458*$$\sqrt{1 - 0.5^2}$$ meters, so TB - TA = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(1c - 0.5c) = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(299,792,458 meters/second)*(0.5) = $$(\sqrt{1 - 0.5^2})/0.5$$*(1 second) = 1.73205081 seconds. Similarly, T'A - TB = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(1c + 0.5c) = (299,792,458 meters)*$$\sqrt{1 - 0.5^2}$$/(299,792,458 meters/second)*(1.5) = $$(\sqrt{1 - 0.5^2})/1.5$$*(1 second) = 0.577350269 seconds.

are those values in the "rest frame" or in the stick?

what are the values in the other frame (the "rest frame" or the stick)?

pess5 said:
v = 0.5c
gamma = 1.1547

Rest Frame …

TB-TA = T(out) = 1 sec = 1/2(2L/c)
T'A-TB = T(back) = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters (proper length)

Moving frame …

TB-TA = T(out) = 1.732 sec = (L/gamma)[1/(c-v)]
T'A-TB = T(back) = 0.57733 sec = (L/gamma)[1/(c+v)]
L’ = 0.866 lt-sec = 259,620,268 meters = L/gamma (contracted length)

By the Lorentz Transformations ...

Per Rest Frame (by definition) ...

TB-TA = 1 sec = 1/2(2L/c)
T'A-TB = 1 sec = 1/2(2L/c)
L = 1 lt-sec = 299,792,458 meters

Per Moving Frame (per LTs) ...

t(out) = gamma(T+vX/c2) = TB-TA

t(out) = gamma(T+vX/c2)
t(out) = (1.1547)[(1)+(0.5)(1)/(1)2]
t(out) = 1.732 sec​

t(back) = gamma(T-vX/c2) = T'A-TB

t(back) = gamma(T-vX/c2)
t(back) = (1.1547)[(1)-(0.5)(1)/(1)2]
t(back) = 0.5774 sec​

Verification ... Roundtrip

t(out)+t(back) = gamma[T(out)+T(back)]
1.732 sec + 0.5774 sec = (1.1547)[1 sec + 1sec]
2.3094 sec = 2.3094 sec​

x(out) = gamma(X-vT)

x(out) = gamma(X-vT)
x(out) = (1.1547)[(1)-(0.5c)/(1)2]
x(out) = (1.1547)[1-0.5]
x(out) = 1.732 lt-sec​

x(back) = gamma(X+vT)

x(back) = gamma(X+vT)
x(back) = (1.1547)[(1)+(0.5c)/(1)2]
x(back) = (1.1547)[1+0.5]
x(back) = 0.5774 lt-sec​

L' = x’ = (x-vt) ... calculated by the outbound leg

x’ = (x-vt)
x’ = ct-vt
x’ = t(c-v)
x’ = (1.732 sec)(1-0.5)
x’ = (1.732 sec)0.5
x’ = 0.866 lt-sec
L' = x’ = 0.866 lt-sec = L/gamma

L' = x’ = (x+vt) ... calculated by the return leg

x’ = (x+vt)
x’ = ct+vt
x’ = t(c+v)
x’ = (0.5774 sec)(1+0.5)
x’ = (0.5774 sec)1.5
x’ = 0.866 lt-sec
L' = x’ = 0.866 lt-sec = L/gamma

pess

according to your calculations, would the observers moving with the moving stick find that the two clocks were not synchronous? and at the same time would the observers in the "rest frame" declare the clocks to be synchronous?

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myoho.renge.kyo said:
are those values in the "rest frame" or in the stick?

what are the values in the other frame (the "rest frame" or the stick)?

Those were the values in the "rest frame", ie the frame where the stick was moving at velocity v. As I explained earlier, in this frame the time between the light passing A and the light passing B is (TB - TA) (because you defined the clocks at the end of the ruler to be synchronized in this frame and ticking at the correct rate for this frame, meaning they'd be out-of-sync and ticking too fast in the stick's own rest frame), the length of the stick is L' in this frame, and because the stick has moved forward a distance of v*(TB - TA) in the time (TB - TA), the total distance the light covers between passing A and passing B is L' + v*(TB - TA) in this frame. So, (distance as measured in 'rest frame')/(time as measured in 'rest frame') is equal to [L' + v*(TB - TA)]/(TB - TA), or L'/(TB - TA) + v. And of course, distance/time for a light beam in any inertial frame must be c, so c = L'/(TB - TA) + v, which you can rearrange to get (TB - TA) = L'/(c - v).

In the stick's own frame (what you called the 'moving frame'), the stick has length L, and since the stick is at rest this is the distance the light will cover in both directions, and of course (distance in moving frame)/(time in moving frame) must equal c, so the light will take a time of L/c to get from A to B, and likewise it will take L/c to get from B to A, meaning the times will be 1 second in either direction.

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JesseM said:
Those were the values in the "rest frame", ie the frame where the stick was moving at velocity v. As I explained earlier, in this frame the time between the light passing A and the light passing B is (TB - TA) (because you defined the clocks at the end of the ruler to be synchronized in this frame and ticking at the correct rate for this frame, meaning they'd be out-of-sync and ticking too fast in the stick's own rest frame), the length of the stick is L' in this frame, and because the stick has moved forward a distance of v*(TB - TA) in the time (TB - TA), the total distance the light covers between passing A and passing B is L' + v*(TB - TA) in this frame. So, (distance as measured in 'rest frame')/(time as measured in 'rest frame') is equal to [L' + v*(TB - TA)]/(TB - TA), or L'/(TB - TA) + v. And of course, distance/time for a light beam in any inertial frame must be c, so c = L'/(TB - TA) + v, which you can rearrange to get (TB - TA) = L'/(c - v).

In the stick's own frame (what you called the 'moving frame'), the stick has length L, and since the stick is at rest this is the distance the light will cover in both directions, and of course (distance in moving frame)/(time in moving frame) must equal c, so the light will take a time of L/c to get from A to B, and likewise it will take L/c to get from B to A, meaning the times will be 1 second in either direction.

i was thinking more along these lines:

given:

L = 299,792,458 m

v = 0.5*c

1/sqrt(1 - v^2/c^2) = 1.1547

the moving clock at A and the clock in the "rest frame" mark TA = 0 s.

-------------------------------

let the moving clocks measure TB - TA = T = 1 s (299,792,458 m/c).

let the clocks in the "rest frame" measure TB - TA = t

------------------------------

by the lorentz transformations,

T = (t - v*L'/c^2)/1.1547

1.1547*T = t - v*L'/c^2

1.1547*T = t - (0.5*c)*L'/c^2

1.1547*T = t - 0.5*L'/c

t = 1.1547*T + 0.5*L'/c

------------------------------

L = (L' - v*t)/1.1547

1.1547*L = L' - v*t

1.1547*L = L' - 0.5*c*t

0.5*c*t = L' - 1.1547*L

t = (L' - 1.1547*L)/0.5*c

--------------------------------

1.1547*T + 0.5*L'/c = (L' - 1.1547*L)/0.5*c

1.1547*T*(0.5*c) + 0.25*L' = L' - 1.1547*L

L' - 0.25*L' = 1.1547*T*(0.5*c) + 1.1547*L

L'*(1 - 0.25) = 1.1547*(0.5*T*c + L)

L' = 1.1547*(0.5*T*c + L)/0.75

L' = 1.1547*(149,896,229 m + 299,792,458 m)/0.75

L' = 1.1547*449,688,687 m/0.75

L' = 692,340,702.505 m

---------------------------------

t = 1.1547*T + 0.5*L'/c

t = 1.1547 s + 1.1547 s

t = 2.3094 s

----------------------------------

t = (L' - 1.1547*L)/0.5*c

t = (692,340,702.505 m - 346,170,351.253 m)/(149,896,229 m/s)

t = 2.3094 s

--------------------------------------

692,340,702.505 m/(c - v) = 4.6188 s = T

692,340,702.505 m/(c + v) = 1.5396 s = (T'A - T)

we can see that the observers moving with the stick find that the two clocks were not synchronous.

-------------------------------------------

692,340,702.505 m/(c - v) = 2.3094 s = t (realize that v = 0)

692,340,702.505 m/(c + v) = 2.3094 s = (T'A - t) (again realize that v = 0). the mark of T'A by the moving clock at A is 6.1584 s (4.6188 s + 1.5396 s), and the mark of T'A by the clock in the "rest frame" is 4.6188 s (2*2.3094 s).

we can see here that at the same time, the observers in the "rest frame" declare the clocks to be synchronous.

--------------------------------

now i know what A. Einstein meant on p.42 in The Principle of Relativity (thanks to all of you!).

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myoho.renge.kyo said:
according to your calculations, would the observers moving with the moving stick find that the two clocks were not synchronous? and at the same time would the observers in the "rest frame" declare the clocks to be synchronous?

Well, I ran the calculation as though the stick were 1 light-sec long in the stick's proper frame. However, any way you slice it, clocks in sync on the ground (ie rest frame) appear unsynchronized per the moving stick. Clock is sync on the moving stick appear unsynchronized per the ground observers. The offset between clocks stems from the -vx/c^2 portion of the LT, so the offset is gamma(xv/c^2) where x is the contracted separation in clocks per the observer.

pess

myoho.renge.kyo said:
you say "Rest Frame..."

what do you mean?

Indeed, I mistakenly assumed the stick's proper frame as the rest frame. Didn't read your secanrio setup carefully enough.

pess

JesseM said:
myoho.renge.kyo is actually defining "rest frame" and "moving frame" in the opposite way--the "moving frame" is the rest frame of the stick (where its length is 1 light-second), and the "rest frame" is the frame where the stick is moving at velocity v. Also, TB - TA and T'A - TB are defined to be the time intervals in the "rest frame" where the stick is moving, not the time in the stick's own frame. This sort of confusion is why I made this suggestion earlier: Aside from that, your numbers for the times agree with the ones I got in my last post.

yes, I misread the scenario setup. Thanx.

I find it best to refer to the "stick frame" and the "observing frame". Either can be stationary per its own vantage, and either considers itself at rest per its own perspective. Two bodies can be at rest with each other when in inertial motion. But then, all inertial bodies are at rest.

pess

myoho.renge.kyo said:
i was thinking more along these lines:

given:

L = 299,792,458 m

v = 0.5*c

1/sqrt(1 - v^2/c^2) = 1.1547

the moving clock at A and the clock in the "rest frame" mark TA = 0 s.

-------------------------------

let the moving clocks measure TB - TA = T = 1 s (299,792,458 m/c).

let the clocks in the "rest frame" measure TB - TA = t

------------------------------

by the lorentz transformations,

T = (t - v*L'/c^2)/1.1547
I don't know what event you're trying to transform the coordinates of here--why do you have L' in there? In the "rest frame", the position-coordinate of the clock at B will only be L' when t=0, at later times that end of the ruler will have moved. Also, you shouldn't be dividing by 1.1547, you should be multiplying by it--the Lorentz transform equation for time is $$T = \gamma*(t - vx/c^2)$$, with $$\gamma = 1/\sqrt{1 - v^2/c^2}$$, which in this case would be $$1/\sqrt{1 - 0.5^2} = 1.1547$$.

In the "rest frame", the A end of the stick was at position x=0 at time t=0, and the B end of the stick was at position x=L'. Because the stick is moving, the B end's position as a function of time will be x(t) = L' + vt. Meanwhile, the light beam's position as a function of time is x(t) = ct, so the light beam catches up with the B end when L' + vt = ct, or when t = L'/(c - v). With L=1 light-second and v=0.5c (and L' = L*0.866 = 0.866 light-seconds), this works out to t=1.732 seconds. Since this is along the path of the light beam, the position of this event must be x=1.732 light-seconds. You can then plug that into the Lorentz transform T = 1.1547*(1.732 s - (0.5 ls/s *1.732 ls)/(1 ls^2/s^2)) = 1.1547*(1.732 s - 0.5*1.732 s) = 1 s.

Alternately, you could figure out the coordinates of the event of the light reaching B in the "moving frame" where the stick is at rest--in this frame, B is always at position X=L, so naturally the light beam reaches B at T=L/c. Then just plug that into the reversed version of the Lorentz transform:

$$x = \gamma * (X + vT)$$
$$t = \gamma * (T + vX/c^2)$$

With X=L and T=L/c, this gives t=(1.1547)*(1.5L/c) = 1.732 seconds, and x=(1.1547)*(1.5L) = 1.732 light-seconds, same as above.

Anyway, the rest of your post is wrong because your initial equation doesn't make sense, but if you think about precisely what event you want to transform, and what coordinates this event should have in either coordinate system, then you should be able to use the Lorentz transform to find the coordinates of the same event in the other coordinate system.

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myoho.renge.kyo said:
let TA = 0 s also be the definite time at which L' is ascertained.
When L' is "ascertained" is irrelevant, L' is just the length of the stick in the "rest frame", which is the same at all times in that frame. What's important is that the stick is moving in the rest frame, so that even though it has a constant length L', the position of the end of the stick is changing over time. So, the position of the B-end at the moment the light reaches it is different than the position the B-end was at when the light was passing the A-end.

Suppose I walk from one end of a train car to another, and the train car is moving down the tracks at high speed in your frame. Wouldn't you agree that the distance between my position when I was at the back end of the car and my position when I reached the front end of the car would be much larger than the length of the car itself in your frame, since the train car would have been moving forward while I was walking along it? It's exactly the same principle with the light beam moving from one end of the stick to the other in a frame where the stick is moving, the distance the light covers in that frame will not be equal to the length of the stick in that frame.

JesseM said:
When L' is "ascertained" is irrelevant, L' is just the length of the stick in the "rest frame", which is the same at all times in that frame. What's important is that the stick is moving in the rest frame, so that even though it has a constant length L', the position of the end of the stick is changing over time. So, the position of the B-end at the moment the light reaches it is different than the position the B-end was at when the light was passing the A-end.

Suppose I walk from one end of a train car to another, and the train car is moving down the tracks at high speed in your frame. Wouldn't you agree that the distance between my position when I was at the back end of the car and my position when I reached the front end of the car would be much larger than the length of the car itself in your frame, since the train car would have been moving forward while I was walking along it? It's exactly the same principle with the light beam moving from one end of the stick to the other in a frame where the stick is moving, the distance the light covers in that frame will not be equal to the length of the stick in that frame.

it is important to realize that L' is ascertained in the following way:

at TA = 0 s, the end A of the stick is located on the x-axis of the "rest frame", and at TB = L'/(c - v) (where v = 0 in the "rest frame", or v = 0.5*c in the stick), the B end of the stick is located on the x-axis of the "rest frame".

pess5 said:
Well, I ran the calculation as though the stick were 1 light-sec long in the stick's proper frame. However, any way you slice it, clocks in sync on the ground (ie rest frame) appear unsynchronized per the moving stick. Clock is sync on the moving stick appear unsynchronized per the ground observers. The offset between clocks stems from the -vx/c^2 portion of the LT, so the offset is gamma(xv/c^2) where x is the contracted separation in clocks per the observer.

pess

L'/(c - v) = 2.3094 s (time of light from A to B. realize that L' is stationary in the "rest frame", v = 0)

L'/ (c - v) = 2.3094 s (time of light from B to A. again, realize that L' is stationary in the "rest frame", v = 0)

therefore, the observers in the "rest frame" would declare the clocks to be in sync.

L'/(c - v) = 4.6188 s (time of light from A to B. here realize that the stick moves with speed v = 0.5*c relatively to L').

L'/(c + v) = 1.5396 s (the time of light from B to A. realize here again that the stick moves with speed v = 0.5*c relatively to L')

therefore, the observers in the stick would find the two clocks were not in sync.

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myoho.renge.kyo said:
it is important to realize that L' is ascertained in the following way:

at TA = 0 s, the end A of the stick is located on the x-axis of the "rest frame", and at TB = L'/(c - v) (where v = 0 in the "rest frame", or v = 0.5*c in the stick), the B end of the stick is located on the x-axis of the "rest frame".
That doesn't make sense, in relativity "length", by definition, is the difference in position between the front end of the object and the back end at a single moment in time in whatever frame you're using. You can't measure the position of the A end at one time and the position of the B end at a different time and call the difference the "length"!

Anyway, I don't know what source you got the equation (TB - TA) = L'/(c - v) from, but I would assume the author derived this equation from the assumption that at time t = TB, the B end is at position x = L' + v*(TB - TA), which means that at time t=TA the B end was at position x=L' (since during the time between TA and TB the B end would move a distance of v*(TB - TA)). And since at t=TA the A end was at position x=0, that means author was indeed assuming that the length L' is ascertained by looking at the difference between the two positions at a single moment in time (namely the moment t=TA, when the A end is at x=0 and the B end is at x=L').

Also, I don't understand what you mean by "located on the x-axis of the rest frame". Aren't we assuming the stick is moving along the x-axis, so that A and B are always at some point on the x-axis? That at least is what I have been assuming in all my derivations, and the equations for the Lorentz transform that we've both been using are based on the assumption that each frame is moving along the other's x-axis. The important thing is where the A and B ends are along the x-axis, namely what their x-coordinates are at different times. What do you think the x-coordinate of the A end is at time TA, and at time TB? What do you think the x-coordinate of the B end is at time TA and time TB?

JesseM said:
I don't know what event you're trying to transform the coordinates of here--why do you have L' in there? In the "rest frame", the position-coordinate of the clock at B will only be L' when t=0, at later times that end of the ruler will have moved. Also, you shouldn't be dividing by 1.1547, you should be multiplying by it--the Lorentz transform equation for time is $$T = \gamma*(t - vx/c^2)$$, with $$\gamma = 1/\sqrt{1 - v^2/c^2}$$, which in this case would be $$1/\sqrt{1 - 0.5^2} = 1.1547$$.

In the "rest frame", the A end of the stick was at position x=0 at time t=0, and the B end of the stick was at position x=L'. Because the stick is moving, the B end's position as a function of time will be x(t) = L' + vt. Meanwhile, the light beam's position as a function of time is x(t) = ct, so the light beam catches up with the B end when L' + vt = ct, or when t = L'/(c - v). With L=1 light-second and v=0.5c (and L' = L*0.866 = 0.866 light-seconds), this works out to t=1.732 seconds. Since this is along the path of the light beam, the position of this event must be x=1.732 light-seconds. You can then plug that into the Lorentz transform T = 1.1547*(1.732 s - (0.5 ls/s *1.732 ls)/(1 ls^2/s^2)) = 1.1547*(1.732 s - 0.5*1.732 s) = 1 s.

Alternately, you could figure out the coordinates of the event of the light reaching B in the "moving frame" where the stick is at rest--in this frame, B is always at position X=L, so naturally the light beam reaches B at T=L/c. Then just plug that into the reversed version of the Lorentz transform:

$$x = \gamma * (X + vT)$$
$$t = \gamma * (T + vX/c^2)$$

With X=L and T=L/c, this gives t=(1.1547)*(1.5L/c) = 1.732 seconds, and x=(1.1547)*(1.5L) = 1.732 light-seconds, same as above.

Anyway, the rest of your post is wrong because your initial equation doesn't make sense, but if you think about precisely what event you want to transform, and what coordinates this event should have in either coordinate system, then you should be able to use the Lorentz transform to find the coordinates of the same event in the other coordinate system.

yes. for simplicity let's say that at TA = 0 s, x = 0, so that at TB = L'/(c - v) (where L' is stationary in the "rest frame", v = 0, or where the stick moves with velocity v = 0.5*c relatively to L'), x = L'.

T = the time (1 s) that light travel 299,792,458 m (the length L of the stick)

t = the time (2.3094 s) that light travel 692,340,702.505 m (the length L' on the x-axis of the "rest frame")

by the lorentz transformations:

T = (t - v*L'/c^2)/sqrt(1 - v^2/c^2)

L = (L' - v*t)/sqrt(1 - v^2/c^2),

where L and T are the values in the stick (the "moving frame") and L' and t are the values in the "rest frame". keep in mind that the observers moving with the stick measure the values L and T with their clock and meter-stick, and that the observers in the "rest frame" meassure the values L' and t with their clock and meter-stick. in other words, when the stick moves relatively to the origin of the "rest frame" with velocity v, the values L and T (as measured by the observers moving with the stick) correspond to the values L' and t (as measured by the observers in the "rest frame"). that's because "the moving stick (at the epoch L'/(c -v) = t = 2.3094; L' is stationary in the "rest frame", v = 0) may not in geometrical respects be perfectly represented by the same stick at rest in a definite position" (A. Einstein). there is nothing magical about this. at TA = 0 s, the light moves from A to B while the stick also moves with velocity v = 0.5*c relatively to the origin of the "rest frame" for 2.3094 s. L' is the distance from A (at TA = 0 s) to B (at the end of 2.3094 s).

JesseM said:
That doesn't make sense, in relativity "length", by definition, is the difference in position between the front end of the object and the back end at a single moment in time in whatever frame you're using. You can't measure the position of the A end at one time and the position of the B end at a different time and call the difference the "length"!

Anyway, I don't know what source you got the equation (TB - TA) = L'/(c - v) from, but I would assume the author derived this equation from the assumption that at time t = TB, the B end is at position x = L' + v*(TB - TA), which means that at time t=TA the B end was at position x=L' (since during the time between TA and TB the B end would move a distance of v*(TB - TA)). And since at t=TA the A end was at position x=0, that means author was indeed assuming that the length L' is ascertained by looking at the difference between the two positions at a single moment in time (namely the moment t=TA, when the A end is at x=0 and the B end is at x=L').

Also, I don't understand what you mean by "located on the x-axis of the rest frame". Aren't we assuming the stick is moving along the x-axis, so that A and B are always at some point on the x-axis? That at least is what I have been assuming in all my derivations, and the equations for the Lorentz transform that we've both been using are based on the assumption that each frame is moving along the other's x-axis. The important thing is where the A and B ends are along the x-axis, namely what their x-coordinates are at different times. What do you think the x-coordinate of the A end is at time TA, and at time TB? What do you think the x-coordinate of the B end is at time TA and time TB?

imagine that the x-axis of the "rest frame" is covered with white paper. now at TA = 0 s, the light departs from A and simultaneously an electrical spark at the base of the stick marks on the paper where the end A of the stick is located. then 2.3094 seconds later, the light is reflected at B and simultaneously another electrical spark marks on the paper where the end B of the stick is located. the length between the two marks on the paper is L'.

myoho.renge.kyo said:
yes. for simplicity let's say that at TA = 0 s, x = 0, so that at TB = L'/(c - v) (where L' is stationary in the "rest frame", v = 0, or where the stick moves with velocity v = 0.5*c relatively to L'), x = L'.
Wait, what do you mean "L' is stationary in the rest frame"? I thought L' stood for the length of the moving stick in the rest frame, while L stood for the length of the same stick in the moving frame (the frame where the stick is at rest). I don't remember you introducing any other objects into this scenario besides the stick. If L' is not the length of the stick as measured in the rest frame, what is it the length of?
myoho.renge.kyo said:
T = the time (1 s) that light travel 299,792,458 m (the length L of the stick)

t = the time (2.3094 s) that light travel 692,340,702.505 m (the length L' on the x-axis of the "rest frame")
But these don't represent the same event! In 1 s in the "moving frame", the light is at the B-end of the stick, correct? But after 2.3094 s in the "rest frame", the light is not at the B-end of the stick, I've already shown that this happens after 1.732 seconds in the rest frame. By 2.3094 s, the light will have already moved well past the B-end of the moving stick.
myoho.renge.kyo said:
by the lorentz transformations:

T = (t - v*L'/c^2)/sqrt(1 - v^2/c^2)

L = (L' - v*t)/sqrt(1 - v^2/c^2),

where L and T are the values in the stick (the "moving frame") and L' and t are the values in the "rest frame".
The values of what? Do you understand that the Lorentz transform shows you the relation between the coordinates of a single event in two different frames? If so, can you say exactly what event you're using in the above equation? If the event you're interested in is the event of the light passing the B-end of the stick, then the x-coordinate of this event in the "rest frame" would not be L' (unless you are assuming L' represents the length of a separate stick at rest in the rest frame, which has a length of 1.732 light-seconds in the rest frame).

Just check the math in your own equations--if L = 1 light-second, and T = 1 second (the coordinates of the light passing B in the moving frame), and v=0.5c, then the equations above will only work if you assume t=1.732 seconds and L' =1.732 seconds:

(t - v*L'/c^2)/sqrt(1 - v^2/c^2) = (1.732 s - [0.5 ls/s * 1.732 ls]/[1 ls^2/s^2])/sqrt(1 - 0.5^2) = (0.866 s)/(0.866) = 1 s = T

and

(L' - v*t)/sqrt(1 - v^2/c^2) = (1.732 ls - [0.5 ls/s * 1.732 s])/sqrt(1 - 0.5^2) = (0.866 ls)/(0.866) = 1 ls = L

But again, this only makes sense if you treat L' as the length of a separate stick at rest in the rest frame, with length 1.732 ls...in that case the B-end of the moving stick will line up with the end of this second stick after 1.732 s in the rest frame, and that will be the moment the light beam passes the ends of both sticks as well.

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JesseM said:
Wait, what do you mean "L' is stationary in the rest frame"? I thought L' stood for the length of the moving stick in the rest frame, while L stood for the length of the same stick in the moving frame (the frame where the stick is at rest). I don't remember you introducing any other objects into this scenario besides the stick. If L' is not the length of the stick as measured in the rest frame, what is it the length of? But these don't represent the same event! In 1 s in the "moving frame", the light is at the B-end of the stick, correct? But after 2.3094 s in the "rest frame", the light is not at the B-end of the stick, I've already shown that this happens after 1.732 seconds in the rest frame. By 2.3094 s, the light will have already moved well past the B-end of the moving stick. The values of what? Do you understand that the Lorentz transform shows you the relation between the coordinates of a single event in two different frames? If so, can you say exactly what event you're using in the above equation? If the event you're interested in is the event of the light passing the B-end of the stick, then the x-coordinate of this event in the "rest frame" would not be L' (unless you are assuming L' represents the length of a separate stick at rest in the rest frame, which has a length of 1.732 light-seconds in the rest frame).

Just check the math in your own equations--if L = 1 light-second, and T = 1 second (the coordinates of the light passing B in the moving frame), and v=0.5c, then the equations above will only work if you assume t=1.732 seconds and L' =1.732 seconds:

(t - v*L'/c^2)/sqrt(1 - v^2/c^2) = (1.732 s - [0.5 ls/s * 1.732 ls]/[1 ls^2/s^2])/sqrt(1 - 0.5^2) = (0.866 s)/(0.866) = 1 s = T

and

(L' - v*t)/sqrt(1 - v^2/c^2) = (1.732 ls - [0.5 ls/s * 1.732 s])/sqrt(1 - 0.5^2) = (0.866 ls)/(0.866) = 1 ls = L

But again, this only makes sense if you treat L' as the length of a separate stick at rest in the rest frame, with length 1.732 ls...in that case the B-end of the moving stick will line up with the end of this second stick after 1.732 s in the rest frame, and that will be the moment the light beam passes the ends of both sticks as well.

i understand your frustration because i experinced it too while trying to understand this. the single event is the motion of light from the end A of the stick to the end B of the stick. if the stick is stationary in the "rest frame", then the observer in the stick measure the value L and T with their clock and meter stick, and the observer on the "rest frame" measure the value L' and t with their clock and meter stick. because the stick does not move relatively to the "rest frame", L = L' and T = t. on the other hand, if the stick moves with velocity v relatively to the origin of the "rest frame", then L = (L' - v*t)/sqrt(1 - v^2/c^2) and T = (t - v*L'/c^2)/sqrt(1 - v^2/c^2). since L' is the distance between two points on the x-axis of the "rest frame", this distance L' is stationary relatively to the "rest frame". therefore, L'/(c - v) = 2.3094 s (since v = 0, in other words, L' is fixed on the x-axis of the "rest frame")

myoho.renge.kyo said:
i understand your frustration because i experinced it too while trying to understand this. the single event is the motion of light from the end A of the stick to the end B of the stick.
You're not understanding what "event" means in relativity, it is something that happens at a single point in space and a single point in time, not a process over an extended time period like light moving from one place to another. In terms of a spacetime diagram, an "event" is always a single point in spacetime, not an extended section of a worldline.
myoho.renge.kyo said:
if the stick is stationary in the "rest frame", then the observer in the stick measure the value L and T with their clock and meter stick, and the observer on the "rest frame" measure the value L' and t with their clock and meter stick. because the stick does not move relatively to the "rest frame", L = L' and T = t. on the other hand, if the stick moves with velocity v relatively to the origin of the "rest frame", then L = (L' - v*t)/sqrt(1 - v^2/c^2) and T = (t - v*L'/c^2)/sqrt(1 - v^2/c^2).
No, that is not the way the Lorentz transformation works! You don't use the Lorentz transformation formula to transform the length of the stick in one frame into the length of the same stick in another! The length transformation formula would just be L' = L*sqrt(1 - v^2/c^2). The Lorentz transformation formula transforms the x and t coordinates of an event in one frame into the X and T coordinate of the same event in another frame. And again, in relativity "event" is something that happens at a single point in spacetime, like the instantaneously brief moment that the light beam passes by the B-end of the moving stick.
myoho.renge.kyo said:
since L' is the distance between two points on the x-axis of the "rest frame", this distance L' is stationary relatively to the "rest frame". therefore, L'/(c - v) = 2.3094 s (since v = 0, in other words, L' is fixed on the x-axis of the "rest frame")
No, you're getting the intended meaning of these formulas completely confused here. The formula L'/(c - v) = (TB - TA) is based on the assumption that L' represents the length of the stick in the "rest frame", not the x-coordinate of the event of the light passing the B-end of the stick in the rest frame. (I assume you copied this formula from somewhere rather than deriving it yourself? I've already explained several times how to derive it if you assume L' represents the length of the stick as measured in the 'rest frame'.) Thus, L' = L*sqrt(1 - v^2/c^2) according to the length contraction formula, giving L' = 0.866 light-seconds. so L'/(c - v) = 0.866 ls /(0.5 ls/s) = 1.732 s, which is exactly the value for the time between the light passing A and the light passing B that both I and pess5 got.

myoho.renge.kyo said:
i understand your frustration because i experinced it too while trying to understand this. the single event is the motion of light from the end A of the stick to the end B of the stick.
That's not a single event, but multiple events. (An "event" is something that happens at a specific place and time.) Here are the relevant events:
(1) The light is emitted at end A
(2) The light arrives (and is reflected) at end B
(3) The light arrives at end A

Each of these three events can be observed and measured by observers in any frame; each frame will assign its own values of position and time for each event according to that frames clocks and coordinate system. Assuming that clocks in each frame are synchronized in the usual way, the Lorentz transformations will allow you to translate between measurements made (of the same event) in two different frames.

Doc Al said:
That's not a single event, but multiple events. (An "event" is something that happens at a specific place and time.) Here are the relevant events:
(1) The light is emitted at end A
(2) The light arrives (and is reflected) at end B
(3) The light arrives at end A

Each of these three events can be observed and measured by observers in any frame; each frame will assign its own values of position and time for each event according to that frames clocks and coordinate system. Assuming that clocks in each frame are synchronized in the usual way, the Lorentz transformations will allow you to translate between measurements made (of the same event) in two different frames.

i see were i am having difficulties here. let me try it this way: the stick is stationary in the "rest frame". the axis of the stick lies along the x-axis of the "rest frame". the end A of the stick coincides with the origin of the "rest frame". L is the distance between the end A of the stick and the end B of the stick (the length of the stick). the length L is measured by a meterstick which is also stationary. a constant velocity v relatively to the origin of the "rest frame" in the direction of increasing x is imparted to the stick. the observer moving together with the meterstick and the stick to be measured, measures the length of the stick again directly by superposing the meterstick, in just the the same way as if all three were at rest. the length of the stick obtained by this operation is also the length L of the stick.

clocks are set up in the "rest frame". these clocks are in sync. a clock is attached to the end A of the stick. another clock is attached to the end B of the stick. the clock attached to the end A of the stick and the clock attached to the end B of the stick are in sync with the clocks set up in the "rest frame", that is to say that their indications correspond at any instant to the "time of the 'rest frame'" at the places where they happened to be. these clocks are therefore in sync in the "rest frame".

a ray of light departs from the end A of the stick at the time TA, and at the time TA the end A of the stick coincides with a point x1 on the x-axis of the "rest frame". (keep in mind that TA is the position of the hands of the clock set up at the point x1 on the x-axis of the "rest frame" and the position of the hands of the clock attached to the end A of the stick.) the ray of light is reflected at the end B of the stick at the time TB , and at the time TB the end B of the stick coincides with the point x2 on the x-axis of the "rest frame". (keep in mind that TB is the position of the hands of the clock set up at the point x2 on the x-axis of the "rest frame" and the position of the hands of the clock attached to the end B of the stick.) the ray of light arrives at the end A again of the stick at the time T'A (the position of the clock attached to the end A of the stick).

the two clocks are in sync if TB - TA = T'A - TB.

L' is the distance between the points x1 and x2 on the x-axis of the "rest frame".

taking into consideration the principle of the constancy of the velocity of light we find that

L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA.

what is the value L' and TB - TA, such that the observers in the "rest frame" would declare the clocks to be in sync if TA is 1 s, the length L is 299,792,458 m, and v = 0.5*c?

and what is the value TB - TA, such that the observers moving with the stick would find the two clocks were not in syn if TA is 0 s, the length L is 299,792,458 m, and v = 0.5*c? thanks!

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myoho.renge.kyo said:
i understand. let me try it this way: the stick is stationary in the "rest frame". the axis of the stick lies along the x-axis of the "rest frame". the end A of the stick coincides with the origin of the "rest frame". L is the distance between the end A of the stick and the end B of the stick (the length of the stick). the length L is measured by a meterstick which is also stationary. a constant velocity v relatively to the origin of the "rest frame" in the direction of increasing x is imparted to the stick at TA (i will explain later what i mean by TA). the observer moving together with the meterstick and the stick to be measured, measures the length of the stick again directly by superposing the meterstick, in just the the same way as if all three were at rest.

clocks are set up in the "rest frame". these clocks are in sync. a clock is attached to the end A of the stick. another clock is attached to the end B of the stick. the clock attached to the end A of the stick and the clock attached to the end B of the stick are in sync with the clocks set up in the "rest frame", that is to say that their indications correspond at any instant to the "time of the 'rest frame'" at the places where they happened to be. these clocks are therefore in sync in the "rest frame".

a ray of light departs from the end A of the stick at the time TA (keep in mind that at the time TA the end A of the stick coincides with the origin of the "rest frame"). (keep in mind also that TA is the position of the hands of the clock set up at the origin of the "rest frame" and the position of the clock attached to the end A of the stick.) the ray of light is reflected at the end B of the stick at the time TB , and at the time TB the end B of the stick coincides with the point x1 on the x-axis of the "rest frame".
We are still assuming the stick is moving, correct? i.e. the B end of the stick was only at x1 at time TB, at earlier or later times (including the time TA) it would be at a different point on the x-axis?
myoho.renge.kyo said:
(keep in mind that TB is the position of the hands of the clock set up at the point x1 in the x-axis of the "rest frame" and the position of the hands of the clock attached to the end B of the stick.) the ray of light arrives at the end A again of the stick at the time T'A (the position of the clock attached to the end A of the stick).
OK.
myoho.renge.kyo said:
the two clocks are in sync if TB - TA = T'A - TB.
This is incorrect. You said earlier that the stick was moving in the "rest frame", but that the clocks at the A end and the B end would match the time in the rest frame. But in the rest frame, when the light departs from the A end, the B end is moving away from the starting point of the light, whereas when the light is reflected from the B end, the A end is moving towards the point where the light was reflected. Since the rest frame says the light itself moves at c in both directions, it should take more time for the light to go from A to B than it does for the light to go from B back to A.
myoho.renge.kyo said:
L' is the distance between the origin and the point x1 on the x-axis of the "rest frame".
OK, x1 is the point on the x-axis where the B-end of the stick is at time TB. And you said earlier that at time TA, the A-end of the stick was at the origin x=0. Since the stick is moving, you're clear that according to this definition L' would not be the length of the stick in the "rest frame", correct?
myoho.renge.kyo said:
taking into consideration the principle of the constancy of the velocity of light we find that

L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA.
False, given your definition of L'. Where did you get this formula? If you derived it yourself, can you explain the derivation? And if you just copied it from somewhere, I feel confident that the author, whoever they were, intended the symbol L' to represent the length of the stick in the "rest frame", which would be equal to the position of the B-end at TA (assuming that the A-end was at the origin at TA), not to the position of the B-end at TB. Again, if you assume the B-end is at position x = L' at time TA, and it's moving along the x-axis at velocity v, then at time TB it will have moved an additional distance of v*(TB - TA) along the x-axis, so its new position will be x = L' + v*(TB - TA). And if the light beam started at the origin at time TA, and reached the B end of the stick at time TB, then we have [distance]/[time] = c which gives:
[L' + v*(TB - TA)] / [(TB - TA)] = c
L'/(TB - TA) + v = c
L'/(TB - TA) = c - v
(TB - TA)/L' = 1/(c - v)
TB - TA = L'/(c - v)

So you can see it is easy to derive the formula you give if we assume L' represents the length of the stick in the "rest frame", so that if the A-end was at x=0 at time TA, the B-end would be at x=L' at time TA, and would later be at x=L' + v*(TB - TA) at time TB.

On the other hand, if you assume that the B-end was at position x=L' at time TB, so that L' is not the length of the stick, then I have no idea how you would derive the formula L'/(c - v) = TB - TA. If you think it is possible that whoever wrote this formula meant L' in the way that you mean it, can you show the derivation of the formula?

Also, note that the two formulas L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA contradict your earlier assertion that "the two clocks are in sync if TB - TA = T'A - TB." TB - TA cannot equal T'A - TB according to these formulas, because L'/(c - v) will not equal L'/(c + v) for any nonzero value of v.

JesseM said:
We are still assuming the stick is moving, correct? i.e. the B end of the stick was only at x1 at time TB, at earlier or later times (including the time TA) it would be at a different point on the x-axis? OK. This is incorrect. You said earlier that the stick was moving in the "rest frame", but that the clocks at the A end and the B end would match the time in the rest frame. But in the rest frame, when the light departs from the A end, the B end is moving away from the starting point of the light, whereas when the light is reflected from the B end, the A end is moving towards the point where the light was reflected. Since the rest frame says the light itself moves at c in both directions, it should take more time for the light to go from A to B than it does for the light to go from B back to A. OK, x1 is the point on the x-axis where the B-end of the stick is at time TB. And you said earlier that at time TA, the A-end of the stick was at the origin x=0. Since the stick is moving, you're clear that according to this definition L' would not be the length of the stick in the "rest frame", correct? False, given your definition of L'. Where did you get this formula? If you derived it yourself, can you explain the derivation? And if you just copied it from somewhere, I feel confident that the author, whoever they were, intended the symbol L' to represent the length of the stick in the "rest frame", which would be equal to the position of the B-end at TA (assuming that the A-end was at the origin at TA), not to the position of the B-end at TB. Again, if you assume the B-end is at position x = L' at time TA, and it's moving along the x-axis at velocity v, then at time TB it will have moved an additional distance of v*(TB - TA) along the x-axis, so its new position will be x = L' + v*(TB - TA). And if the light beam started at the origin at time TA, and reached the B end of the stick at time TB, then we have [distance]/[time] = c which gives:
[L' + v*(TB - TA)] / [(TB - TA)] = c
L'/(TB - TA) + v = c
L'/(TB - TA) = c - v
(TB - TA)/L' = 1/(c - v)
TB - TA = L'/(c - v)

So you can see it is easy to derive the formula you give if we assume L' represents the length of the stick in the "rest frame", so that if the A-end was at x=0 at time TA, the B-end would be at x=L' at time TA, and would later be at x=L' + v*(TB - TA) at time TB.

On the other hand, if you assume that the B-end was at position x=L' at time TB, so that L' is not the length of the stick, then I have no idea how you would derive the formula L'/(c - v) = TB - TA. If you think it is possible that whoever wrote this formula meant L' in the way that you mean it, can you show the derivation of the formula?

Also, note that the two formulas L'/(c - v) = TB - TA and L'/(c + v) = T'A - TA contradict your earlier assertion that "the two clocks are in sync if TB - TA = T'A - TB." TB - TA cannot equal T'A - TB according to these formulas, because L'/(c - v) will not equal L'/(c + v) for any nonzero value of v.

i am sorry. i understand now where i am having difficulties. i will set up the problem differently in the near future in a different thread. thanks!

## 1. How does relativity affect the length of objects?

The theory of relativity states that the length of an object can appear to change depending on the observer's frame of reference. This is known as length contraction, where an object in motion will appear shorter in the direction of motion when measured by an observer at rest.

## 2. Can time really slow down or speed up?

According to the theory of relativity, time is not absolute and can appear to slow down or speed up depending on the observer's frame of reference. This is known as time dilation, where time appears to pass slower for an object in motion compared to an object at rest.

## 3. How does gravity affect the relativity of lengths and times?

Einstein's theory of general relativity explains that gravity is not a force, but rather a curvature of spacetime. This means that gravity can affect the relativity of lengths and times, causing objects to appear shorter and time to pass slower in the presence of a strong gravitational field.

## 4. Is there any evidence to support the theory of relativity?

There is a significant amount of evidence to support the theory of relativity, including the observations of the bending of starlight by the sun's gravitational field, the measurement of time dilation in high-speed particle accelerators, and the accurate predictions of the orbits of planets and satellites.

## 5. Can the average person understand the concept of relativity?

While the theory of relativity may seem complex, the basic concepts can be understood by the average person with some effort and explanation. Many popular science books and documentaries have been created to help explain the theory in a more accessible way.

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